imgur.com/BvYHB6z
Is there a way to find h ? I mean, yes I could split the big triangle in two right angle triangles and then solve for h/2 that would be equal to sine (45/2) * 2 if I am not wrong. But I gotta do it without using trigonometry. So I wonder if there is a way do it. If you know how, pls help me mens.
btw, A is a point in the coordenates (0,0) and B is in (1,0). The lenght of the arc B B' is 0.785 if im not wrong.

0,0 to 0,1 is 10 squares aka 0.1s, just count and use logic lmao
edit: the first 'h' on the left has a slope of 1, the 2nd 'h' which you're trying to find has a slope of -2 aka h is 0.5 because its exactly half of the left side which is (1)

no its not exactly half nvm but its along those lines which i cant be fucked to do since i left school 2 years ago lmao. basically just find the slopes using logic and make an equation. the equation is some shit like point A (x,y) * 2(m aka slope) or some shit like that, do it for the left triangle and then equate it to the right triangle and solve for the right triangle's "h"

What's the entire question? The triangle is inscribed in a circle. You can find the angles using the vertical line as a reference to create 2 triangles, so you know you have two 45 degree angles and a 90 degree angle. You can also just use the grid in the back to know how many units long the line is. If it doesn't give you any units or any measurements to work with though you can't find it's length because it could just be dilated

Well its an isosceles triangle with side lenght 1 with BAB' making an angle of 45 degrees and I should solve for the lenght of its other side.
Using the grid is a good idea but that would be a gross approximation tbh

Anyway, the triangle made by A, B1 and that unspecified point is an isosceles triangle, since y = 90 and A = 45, so B1 is 180-90-45 = 45. Just remember this for any time you spot a right triangle with one angle being 46. So the short, equal sides are of some length, x, and by the Pythagorean theorem, the hypotenuse, which is 1, gives the equation x^2 + x^2 = 1^2, in other words, x = 0.5sqrt(2). AB1 = AB, since A is the middlepoint of a circle and B1 and B are both on it, so j = x = 0.5sqrt(2) and that short length on AB is 1- 0.5sqrt(2). Now use the Pythagorean theorem again.
h^2 = (0.5sqrt(2))^3 + (1-0.5sqrt(2))^2 and you can simplify that.

oh you right ! thank you.
Anyway, I mean the distance of the unspecified point to B is equal to 1 - j.
Do you think there is a way to generalize it ? I mean, if the angle is not 45 in the triangle B'BA it is impossible to solve ?

Because of the angle, it's 45°, if you go 3 squares to the right from your B', then 3 squares up, then count the squares from where you are all the way to the B angle, you get 1
Pythagoras then : sqrt(1+1) = sqrt(2)

using Thales the line below is 0,70710678118654752440084436210485 for the first triangle and then 0,29289321881345247559915563789515‬ for the second triangle, if add them it's = 1 to prove that it is correct
we can use pythagoras to find the "j" line now by using : 1 = sqrt( j² + 0,7...²)
1² = j² + 0,7...²
j = sqrt (1² - 0,7...²)
j = 0,70710678118654752440084436210485
j is the same as the line below that we found
you can now search for "h" using pythagoras once again
h = sqrt (0,7...² + 0,2...²)
h = 0,7653668647301795434569199680608‬

this is correct but you dont even need to know about the principles because we know that j is sqrt(2) and the base between A and (0,0) is sqrt(2)
just using pythagorean theorem you get sqrt(sqrt(2)²)+(1-(sqrt(2))²) = 0.765...

Just help yourself with the squares, use pythagorean theorem. I'm sure it would be pretty easy. Do you know the length of a square or something? If so then you can easily solve it.
If 1 is the length of |AB| then then one square is 0,1 and a square inside that square would be 0,02. That's all the info you need to solve it.

So easy that I can't stop laughing at retards on hltv.
Have you heard about something know as similarity and pythagoras theorem?
Let the foot of perpendicular be P
Then triangle AB'P is similar to triangle B'BP
So j^2=x*(1-x) where x is length of AP.
Now find value of j^2 from 2 right triangles and this value.
Equate all 3, factorise them and done.

first answer is without trigonometry because I'm not blind.
I would never produce that solution if there were no restrictions.
Second one is obvious solution and straight forward. Just to show you beauty of maths.